Question

Assuming a lift curve slope of 0.11 per degree angle of attack for a two-dimensional airfoil, calculate the lift coefficients at 2, 6 and 10 degrees angle of attack for:

a. A symmetric airfoil.
b. A positively cambered airfoil with a zero lift angle of attack of -1.3 degrees.
c. A reflexed airfoil with a zero lift angle of attack of 1.75 degrees.

Answer 1

a) at 2° => 0.22

6° => 0.66

10° => 1.1

b) at 2° => 0.363

6°=>0.803

10° => 1.243

c) at 2° => 0.0275

6° => 0.4675

10° => 0.9075

Step-by-step explanation:

We are given:

Lift curve slope = 0.11 per degree

`a_1=2degrees; a_2=6degrees; a_3=10degrees`

a) to find symmetric airfoil, we use the formula;

`C_L = C_ha * a`

`at a_1 = 2 => C_L = 0.11*5`

= 0.22

`at a_2 = 6=> C_L = 0.11*6`

=0.66

`at a_3 =10=> C_L = 0.11*10`

= 1.1

b) to find cambered airfoil, with a zero lift angle of attack at -1.3°, we use:

`(C_L-0)/(a+1.3) = 0.11`

Making C_L subject of the formula, we have:

`C_L = 0.11(a +1.3)`

At a = 2 => 0.11(2+1.3) = 0.363

At a = 6 => 0.11(6+1.3) = 0.803

At a = 10 => 0.11(10+1.3)= 1.243

c) to calculate for a eflexed airfoil with a zero lift angle of attack of 1.75 degrees, we use:

`(C_L-0)/(a-1.75)`

`C_L = 0.11(a-1.75)`

At a= 2 => 0.11(2-1.75) = 0.0275

At a = 6 => 0.11(6-1.75) = 0.4675

At a = 10 => 0.11(10-1.75) = 0.9075